2742

2742. Painting the Walls (Hard)

You are given two 0-indexed integer arrays, cost and time, of size n representing the costs and the time taken to paint n different walls respectively. There are two painters available:

• A paid painter that paints the ith wall in time[i] units of time and takes cost[i] units of money.
• A free painter that paints any wall in 1 unit of time at a cost of 0. But the free painter can only be used if the paid painter is already occupied.

Return the minimum amount of money required to paint the n walls.

 

Example 1:

Input: cost = [1,2,3,2], time = [1,2,3,2]
Output: 3
Explanation: The walls at index 0 and 1 will be painted by the paid painter, and it will take 3 units of time; meanwhile, the free painter will paint the walls at index 2 and 3, free of cost in 2 units of time. Thus, the total cost is 1 + 2 = 3.

Example 2:

Input: cost = [2,3,4,2], time = [1,1,1,1]
Output: 4
Explanation: The walls at index 0 and 3 will be painted by the paid painter, and it will take 2 units of time; meanwhile, the free painter will paint the walls at index 1 and 2, free of cost in 2 units of time. Thus, the total cost is 2 + 2 = 4.

 

Constraints:

• 1 <= cost.length <= 500
• cost.length == time.length
• 1 <= cost[i] <= 106
• 1 <= time[i] <= 500

Companies: Amazon, Adobe, Snowflake, Apple

Related Topics:
Array, Dynamic Programming

Hints:

• Can we break the problem down into smaller subproblems and use DP?
• Paid painters will be used for a maximum of N/2 units of time. There is no need to use paid painter for a time greater than this.

Solution 1.

Let dp[k] be the minimum cost to finish k walls.

Initially

dp[0] = 0
dp[k>0] = inf

For ith wall, we update dp[j] with this dp equation

dp[j] = min(dp[j], dp[max(j - time[i] - 1, 0)] + cost[i])

If we paint ith wall, we increase cost by cost[i], and we can paint time[i] more walls with free painters. So, we can paint time[i] + 1 walls with this choice. Thus, we can try to use dp[max(j - time[i] - 1, 0)] + cost[i] to update dp[j].

// OJ: https://leetcode.com/problems/painting-the-walls
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N)
// Ref: https://leetcode.com/problems/painting-the-walls/solutions/3650707/java-c-python-7-lines-knapsack-dp/
class Solution {
public:
    int paintWalls(vector<int>& cost, vector<int>& time) {
        int n = cost.size();
        vector<int> dp(n + 1, 1e9);
        dp[0] = 0;
        for (int i = 0; i < n; ++i)
            for (int j = n; j > 0; --j)
                dp[j] = min(dp[j], dp[max(j - time[i] - 1, 0)] + cost[i]);
        return dp[n];
    }
};